Find $\lim_{x\to\infty}\dfrac{2x+\sin(x)}{x+7}$. Choose 1 answer: Choose 1 answer: (Choice A) A $0$ (Choice B) B $1$ (Choice C) C $2$ (Choice D) D The limit doesn't exist
When dealing with limits that include $\sin(x)$, it's important to remember that $\lim_{x\to\infty}\sin(x)$ doesn't exist, as $\sin(x)$ keeps oscillating between $-1$ and $1$ forever. ${2}$ ${4}$ ${6}$ ${8}$ ${\llap{-}4}$ ${\llap{-}6}$ ${\llap{-}8}$ ${2}$ $y$ $x$ $y=\sin(x)$ This doesn't necessarily mean that our limit doesn't exist. Think what happens to $\dfrac{2x+\sin(x)}{x+7}$ as $x$ increases towards positive infinity. $\sin(x)$ oscillates between $-1$ and $1$. This can be represented mathematically by the following double inequality: $\dfrac{2x-1}{x+7}\leq\dfrac{2x+\sin(x)}{x+7}\leq\dfrac{2x+1}{x+7}$ The result is a graph that's always between the graphs of $y=\dfrac{2x\pm 1}{x+7}$ (the dashed lines). ${5}$ ${10}$ ${15}$ ${\llap{-}5}$ ${\llap{-}10}$ ${\llap{-}15}$ ${5}$ ${10}$ ${15}$ ${\llap{-}5}$ ${\llap{-}10}$ ${\llap{-}15}$ $y$ $x$ Since $\lim_{x\to\infty}\dfrac{2x\pm 1}{x+7}=2$, so must our limit be equal to $2$. In conclusion, $\lim_{x\to\infty}\dfrac{2x+\sin(x)}{x+7}=2$.